A particle moving in the xy plane undergoes a displacement
of s = (2iˆ + 3jˆ) while a constant force F = (5iˆ + 2jˆ)N
acts on the particle. The work done by the force F is
(a) 17 joule (b) 18 joule
(c) 16 joule (d) 15 joule
Answers
Answered by
1
Answr:
W=Fˉ.sˉ=(5i^+2j^).(2i^+3j^)=10+6=16J.
Answered by
0
Answer:
Your answers is 16joule.
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